Answer to Question #183942 in Electricity and Magnetism for KyKy

Question

A particle of charge -6.25 nC experiences a force of 1.25 N when it is moving at 3.75 x 10^-4 m/s. Its motion is perpendicular to the magnetic field produced by a current-carrying wire, which is 3.0 cm from it. (a) What is the magnitude of the magnetic field? (b) How much current flows through the wire?

Answer 1

a) Lorentz force, acting on the particle is $F = q v B$, from where the magnitude of the magnetic field is $B = \frac{F}{q v} \approx 5.3 \cdot 10^{-5} T$.

b) The magnitude of magnetic field, produced by the current-carrying wire at distance $r$ from the wire is $B = \frac{\mu_0 I}{2 \pi r}$, where $\mu_0 = 4 \pi \cdot 10^{-7} T \cdot m / A$ is the permeability of free space. From the last formula, $I = \frac{2 \pi r B}{\mu_0} \approx 7.95 A$.